Question : Two line charts are given below. Line chart 1 shows the ratio of the number of males to the number of females in two companies A and B for the 5 years. Line chart 2 shows the total number of males (both companies A and B) and total number of females (both companies A and B) for the 5 years.


What is the ratio of the number of males in company B in Y1 to the total number of females in company A in Y3 and Y5?

Option 1: 117 : 218

Option 2: 117 : 215

Option 3: 129 : 215

Option 4: 119 : 218

Correct Answer: 117 : 215

Solution : Let the number of males and females in Y1 in company A be $11k$ and $10k$ respectively.
In Y1, the number of males and females is $9a$ and $10a$ in company B.
$11k+9a=21100$ (equation 1)
$10k+10a=20600$ (equation 2)
Solving the above equations, we get,
$a=780$
The number of males in company B in Y1 $=9a=9\times780=7020$.
Let the number of males and females in Y3 in company B be $27x$ and $20x$ respectively. (Since the ratio of male to female is 1.35)
In Y3, the number of males and females is $16p$ and $20p$ in company A. (Since the ratio of male to female is 0.8)
$27x+16p=18025$ (equation 3)
$20x+20p=16000$ (equation 4)
Solving the above equations, we get,
$x=475$ and $p=325$
The number of females in company A in Y3 $=20p=20\times325=6500$.
Let the number of males and females in Y5 in company A be $28q$ and $20q$ respectively.
In Y5, the number of males and females is $17r$ and $20r$ in company B
$28q+17r=13550$ (equation 5)
$20q+20r=11800$ (equation 6)
Solving the above equations, we get,
$q=320$
The number of females in company A in Y5 $=20x=20\times320=6400$
So, the required ratio $=7020:(6500+6400)$
$⇒7020:12900=117:215$
Hence, the correct answer is 117: 215.